 Solution

Let $x_1,\dots ,x_s$ contain all free variables of $M$. The lambda term $M$ is a solvable lambda term if there are closed lambda terms $P_1,\dots ,P_k$ such thatAdditional arguments beyond the first $s$ are allowed. A head normal form is a term $\lambda u_1\cdots u_n.y{M}_1\cdots M_r$ with a variable at the head; its arguments need not be beta-normal.

If $M$ reduces to this head normal form, close its free variables and supply every resulting initial binder with the same closed projection(take $P=I$ when $r=0$). The head variable becomes $P$ and discards all $r$ argument terms, even divergent ones, leaving $I$. Thus a head normal form implies solvability.

Conversely suppose a closure and applications of $M$ yield $I$. The standardization theorem for beta reduction supplies a finite successful head-reduction sequence. Until $M$ exposes a head variable, its head contractions lie within $M$: a closing substitution cannot create a new head redex there, and supplied arguments cannot be used until an initial abstraction is exposed. Each such contraction commutes with the closing capture-avoiding substitution. Project the successful head computation back onto $M$, ignoring contractions in supplied arguments and the consumption of its initial binders. If this projected head computation did not reach a variable-headed spine, the full computation could not reach $I$ either. It therefore gives a finite head reduction of $M$ to a head normal form. ConsequentlyThe same head-trace argument shows that an unsolvable lambda term stays unsolvable under substitution and application. This also justifies the unsolvability conclusion in the preceding solution.